Computing flow of vectors fields with partial derivatives

Question

Define vector fields $X$ and $Y$ on the plane by $$ X = x \dfrac{\partial}{\partial x} - y \dfrac{\partial}{\partial y}, \qquad Y = x \dfrac{\partial}{\partial y} + y \dfrac{\partial}{\partial x}.$$ Compute the flows $\theta , \psi$ of $X$ and $Y$, and verify that the flows do not commute by finding explicit open intervals $J$ and $K$ containing $0$ such that $\theta_s \circ \psi_t$ and $\psi_t \circ \theta_s$ are both defined for all $(s,t) \in J \times K$, but they are unequal for some $(s,t)$.

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To solve for the flow I need to solve the PDE. Every example I've encountered dealt with a simple ODE that could be solved via separation of variables. This makes for a dumb question but I've never taken a PDE course, so how do I solve the PDE? For the second part of the question, once I figure out how to solve the PDE, I think I'll be able to show they don't commute, just figured I'd post the full question.

Answer 1

This is a problem from my book Introduction to Smooth Manifolds (2nd edition). You don't need to solve a PDE -- to compute the flow of a vector field entails solving a system of ODEs. If you have my book handy, Examples 9.1 and 9.8 illustrate how this is done.

Answer 2

I made some confusion on my previous post. I will try to detail the answer of Jacke Lee.

He wrote a very great book where you can find it in detail, so this is just a simple explanation.

You must remember that the flow is a family of integral curves for a specific vector field.

Let $\gamma(t) = (x(t),y(t))$ the candidate for the flow. We must solve

$$X(\gamma(t)) = \gamma'(t).$$

Now the fist question is: How can we make sense of $\gamma'(t)?$ It must be an element of the tangent space at $\gamma(t).$ Now we verify that $$\gamma'(t) = x'(t)\partial_{x(t)} + y'(t)\partial_{y(t)}.$$

On the other side we have $X(\gamma(t)) = x(t)\partial_{x(t)} -y(t)\partial_{y(t)}.$ Once $X(\gamma(t)) = \gamma'(t)$ it assigns us:

$$x'(t) = x(t), y'(t) = -y(t).$$ Now it is easy to compute the solution.